Given random variables $X_1,...,X_n \overset{i.i.d.}{\sim} N(\mu, \sigma^2)$ where the variance $\sigma^2$ is known let the null hypothesis be $H_0: \mu = \mu_0$

For the statistic $T=\sum_{i=0}^nX_i$ derive the distribution of $T$ under $H_0$

Calculate $P_{\alpha}$ the probability of making a type 1 error.

I've said that since $E[X_i]<\infty$ and $var(X_i) = \sigma^2$ the central limit theorem is applicable and $\frac{1}{\sqrt{n}}(T - n\mu) \sim N(0, \sigma^2)$ and so for $n$ "large enough" I could say that $T\overset{H_0}{\sim} N(\mu_0, \sigma^2)$

I'm wondering if this is correct though, especially the last step for the distribution under $H_i$

For the probability of a type 1 error I said $P_{\alpha} = P( T \geq \nu | H_0) = 1 - P(T < \nu | H_0) = 1 - \Phi(\frac{\nu - \mu_0}{\sigma})$

where $\nu$ would depend on the unkown $\alpha$ and $\Phi$ is the cumulative distribution function of the standard normal distribution. But this is a conditional probability and it seems to me I shouldn't just be able to say that this can be done with the cdf, unless the independence of the $X_i$ allows this?


First: $\frac{1}{\sqrt{n}}(T - n\mu) \sim N(0, \sigma^2)$ does not imply $T\overset{H_0}{\sim} N(\mu_0, \sigma^2)$ since $\mu_0$ pertains to the mean of a single $X_i$ not the sum. Instead $\frac{T}{n}\sim N(\mu_0, \frac{\sigma^2}{n})$ under the null hypothesis.

Also, since your data are already normally distributed, you don't need to invoke the CLT or any asymptotic results. $T$ will be exactly a normal distribution.

Now, let's define our hypothesis test as $\mathcal{T}(T|a,b,n)=1$ if $\frac{T}{n} \notin [a,b],\;\; \mathcal{T}=0\; o.w.$

The Type I error rate ($\alpha$) is just the expected value of $\mathcal{T}(T|a,b,n)$ evaluated under the null hypothesis:

$$E[\mathcal{T}(T|a,b,n)|E[T]=\mu_0] = \Phi\left(\sqrt{n}\sigma [a-\mu_0]\right)+\left [1-\Phi \left(\sqrt{n}\sigma[b-\mu_0]\right)\right]$$

  • $\begingroup$ Taking your comments on board I decided to see if I could prove directly that $T$ would have a normal distribution via characteristics function and I got $\frac{T}{\sqrt{n}} \sim N(\sqrt{n}\mu_0, \sigma^2)$. Could you perhaps elaborate how you got your result? Thanks for your help! $\endgroup$ – Longeyes Nov 1 '14 at 8:08
  • $\begingroup$ @Longeyes sorry, made typo early in derivation and then incorporated it into the rest of my post. I've corrected it. $\endgroup$ – user76844 Nov 1 '14 at 11:52

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